I like the gin thing. Maybe that's something I'll mention next time I try to explain to a relative what topology is. :-)
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Saul GlasmanMar 25 '10 at 11:37

12

Nitpick: there doesn't have to be a stationary gin molecule, just one which ends up at the same point where it starts.
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Pete L. ClarkMar 25 '10 at 15:07

3

I got a negative critics of my book Matrices (Springer GTM 216) because I used Brouwer FPT to prove that every $n\times n$ real matrix with non-negative entries has a non-negative eigenvalue (its spectral radius). Yet I intended only to illustrate the powerness of Brouver's FPT. The reviewer took it too seriously.
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Denis SerreJan 24 '11 at 15:36

4

I wonder if this should be community wiki. It's asking for a list of answers after all, and it's unclear if there can be a "best" answer.
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David WhiteSep 22 '11 at 13:24

17 Answers
17

The theorem is equivalent to the determinacy of the Hex game. That's a very famous `application'.

The details can be found in [David Gale (1979). "The Game of Hex and Brouwer Fixed-Point Theorem". The American Mathematical Monthly 86: 818–827], a beautiful paper, which JSTOR serves at http://www.jstor.org/stable/2320146.

Mariano, do you really mean equivalent here? I would interpret that to mean that over a very weak base theory, as in Reverse Mathematics, you can prove Brouwer's theorem just from the assumption that Hex is determined. This seems unlikely for any finite instantiation of Hex, since the determinacy of finite games amounts to De Morgan's rules of logic. Perhaps you are referring to some kind of continuous analogue of Hex? Could you explain?
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Joel David HamkinsMar 25 '10 at 18:20

It is a much less technical sense of 'equivalent', I guess... If you know the truth of one of the statements, the truth of the other follows `easily'.
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Mariano Suárez-Alvarez♦Mar 25 '10 at 19:11

2

"determinacy" is not the right word to use here; the relevant fact is that Hex cannot end in a draw (a "topological" fact; any way to assign half the cells to each player gives at least one player a winning path).
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Sridhar RameshMar 25 '10 at 22:07

2

@Joel: This is one area where reverse mathematics as it is currently set up does not quite capture the informal sense of "equivalent." Many people feel intuitively that Sperner's lemma and Brouwer's fixed-point theorem are "equivalent," in that the "tricky part" is the same and you can pass from one to the other via "straightforward" reasoning. However, Sperner's lemma is provable in $RCA_0$ while Brouwer's fixed-point theorem requires $WKL_0$. Very roughly speaking, you need greater logical strength to take a limit, but of course logical strength isn't the same as psychological difficulty.
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Timothy ChowJan 24 '11 at 16:09

Both Sperner's lemma and the non-draw in Hex are of course easier than Brouwer fixed point theorem since both posess elementary and short arguments.
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Lennart MeierJan 24 '11 at 22:18

The existence of mixed Nash equilibriums in multiplayer games can be proved using the Brouwer fixed point theorem. If you imagine that each players is trying to improve his results based on the current actions of his opponents, then there is some combination of strategies so that each player can't improve.

It has to be noted, though, tha Nash's original proof of the result was based on Kakutani's generalized fixed point theorem.
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J. H. S.Mar 25 '10 at 5:37

5

Well, the first published proof used the Kakutani fixed point theorem, which was suggested to Nash by David Gale. However, he had a proof based on the Brouwer fixed point theorem before and the proof in Nash's thesis uses the Brouwer fixed point theorem.
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Michael GreineckerMar 25 '10 at 9:59

In his Annals paper Nash gives a proof based on Brouwer's fixed point theorem. He says that this proof is a considerable improvement of the previous one published in PNAS, which was based on Kakutani's generalized fixed point theorem.
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GH from MONov 15 '12 at 15:42

Just for fun you could give them the 2-dimensional version of the standing-in-a-train theorem. That theorem says that if you want to go to sleep while standing up in a train that goes along a perfectly straight track, there is some starting angle that will cause you not to fall over. (Proof: if the angle is almost all the way forwards, then you will fall forwards; if it is almost all the way backwards, then you will fall backwards; by the intermediate value theorem there must be an angle that leaves you still standing at the end.) Ian Stewart argues vigorously that the implicit continuity assumption (that your final position depends continuously on your initial position) is wrong, but you can just forget about that.

Now let's suppose that you are standing on a surface that can move horizontally in any direction. (Perhaps you are in a boat, say.) This time if you fall, your head will be somewhere in a circle centre your feet and radius your height. Assuming that the position you end up in depends continuously on your starting position, this defines for you a continuous map of the disc that preserves the boundary. By Brouwer it is not a retraction, so there must be a starting position that stops you falling over.

Even if the continuity assumption is not in the end justified, I think this is a good and amusing illustration of the theorem.

As a bonus: this is a good illustration of the importance of the continuity assumption You're free to believe or disbelieve the existence of this magical position depending on whether you buy the continuity of the "map".
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Thierry ZellJan 24 '11 at 15:58

The Brouwer Theorem can be used to prove that a mapping of ${\bf R}^n$ to itself that
has bounded displacement, in the sense that any point is moved at most a fixed amount from its original location, is onto. This seems be a folklore result. I wonder if anyone has a reference for it.

@7-adic: I'm sure one can prove the the FTA using Brouwer (or Poincaré-Miranda) somehow, but I have yet to see a (correct) proof (there are famous incorrect derivations) - could you supply a reference ? Kind regards, Stephan F. Kroneck.
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Stephan F. KroneckJan 5 '12 at 9:44

The following is not exactly an application, but rather a funny picture to illustrate the theorem, precisely in the form of the non-retraction theorem.

Suppose a shark jumps into a shoal of fish (a kind of big ball). The small fishes start escaping in all directions towards the border of the shoal, where the fishes stand still. Yet they escape with a certain disposition to follow a continuous flow, as they usually do, since everybody tends to follow its neighborhoods. But since there is no continuous retraction to the boundary, somebody doesn't know where to go, and stay there for a moment, much to the shark's satisfaction. There is also a 2D version, with a wolf entering into a herd of sheeps. This is just a funny picture, though I like to think that there is some truth in it.

In many nonlinear equations, the existence of a solution (but not its uniqueness) follows from a topological argument in the vein of BFP Theorem. The BFP is at work especially when the equation is posed in some finite dimensional vector space, and you can establish an a priori estimate of the size of the solution. This means that you know a ball $B_R$ containing all the solutions.

The most important example of this situation is the stationary Navier-Stokes equation, with Dirichlet condition $u=0$ on the boundary of the domain. Of course, the ambient space is infinite dimensional, so you first establish the existence of an approximate solution in a subspace of dimension $n$ (Galerkin procedure); this is where you use the BFP Theorem, or its equivalent form that a continuous vector field over $B_R$ which is outgoing on $\partial B_R$ must vanish somewhere. Then passing to the limit as $n\rightarrow\infty$ is pedestrian.

The BFP Theorem is a consequence of the fact that the Euler-Poincaré characteristic of the ball is non-zero. There are counterparts when you work on a compact manifold (with boundary) whose EPC is non-zero. This happened to me in a very interesting way. I considered the free fall of a rigid body in water filling the entire space. The mathematical problem is a coupling between Navier-Stokes and the Euler equation for the top. I looked at a permanent regime, in which the solid body has a time-independent velocity field, and that of the fluid is time-independent as well, once you consider it in the moving frame attached to the solid. The difficulty is that you don't know a priori the direction of the vertical axis (the direction of gravity) in this frame. After a Galerkin procedure, the problem reduces to the search of a zero of a tangent vector field over $B_R\times S^2$. This vector field is outgoing on the boundary $\partial B_R\times S^2$. Because
$$EP(B_R\times S^2)=EP(B_R)\cdot EP(S^2)=1\cdot2\ne0,$$
such a zero exists. Therefore the permanent regime does exist. Remark that because $EP=2$, we even expect an even number of solutions when counting multiplicities, at least at each level of the Galerkin approximation.

The Schauder fixed point theorem can be proved using the Brouwer fixed point theorem. It says that if $K$ is a convex subset of a Banach space (or more generally: topological vector space) $V$ and $T$ is a continuous map of $K$ into itself such that $T(K)$ is contained in a compact subset of $K$, then $T$ has a fixed point.

The Schauder fixed point theorem in its turn is an important tool for existence proofs in differential equations. One easy application is the Peano existence theorem, but there are also more sophisticated examples.

caveat: The Schauder theorem for arbitrary (hausdorff!) topological vector spaces is hard compared to the case of Banach spaces or even locally convex TVS. The general case was only proven in 2002 by Cauty. But of course the locally convex case is the interesting one for applications.
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Johannes HahnJan 25 '11 at 12:55

One standard consequence of Brouwer's theorem is Borsuk's antipodal mapping theorem, which in turn is used to prove that if $E$, $F$ are subspaces of a normed space and the dimension of $E$ is strictly less than the dimension of $F$, then there is a unit vector in $F$ whose distance to $E$ is one. This result is used often in Banach space theory.

@Bill Johnson: That Borsuk's antipodal theorem implies Brouwer's is folklore (cp. Granas' monograph), but how does one argue in the other direction ? I'd be very interested to see a proof ! Kind regards, Stephan F. Kroneck.
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Stephan F. KroneckJan 5 '12 at 9:40

It is worth noting, however, that this can be proven in several more elementary ways. It is a consequence of linear programming duality / the separating hyperplane theorem, and it can also be proven by just iterating the Markov chain if one takes some care dealing with the case of periodic chains. The Brouwer proof is certainly shorter, though.
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Noah SteinMay 27 '10 at 12:21

Noah, can you (or anyone else) give references for the other proofs? Thanks.
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Phil IsettSep 16 '11 at 17:35

In fact, this proof is flawed, for it applies Brouwer fixed point theorem to a discontinuous function. For a correct proof of the fundamental theorem of algebra using the Brouwer fixed point theorem, see M. K. Fort, Jr, Some properties of continuous functions, Amer. Math. Monthly 59 (1952), p. 372-375.
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ACLFeb 12 '13 at 21:37

I've been thinking about a very similar issue: I was considering giving a talk about the Brouwer fixed point theorem to some math majors (but not necessarily ones very familiar with point-set topology).

There is an elementary proof which uses Sperner's lemma; see Michael Henle's book A Combinatorial Introduction to Topology for details. You can outline a proof of Sperner's lemma pretty quickly (induction on dimension, and dimension 1 is easy), and from that, you can wave your hands (or be more precise, depending on how much your audience knows about compactness) to get Brouwer's theorem. Since Sperner's lemma holds in higher dimensions, you can prove the fixed point theorem in higher dimensions, also.

I gave a similar talk last fall. I did not use sperners lemma though, instead I just computed that $\pi_1(D^2)$ has one element and $\pi_1(S^1)$ has more than that. Then you use functoriality, which I had to explain. I also used $\mathbb{C}-0$ to model $S^1$. It worked reasonably well, and was not very formal.
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Sean TilsonJan 25 '11 at 2:25

Simple and example, for which You may perform demonstration here, is to use some piece of rubber balloon and a lamp.If You stretch it and show that it leaves a shadow on the table. Then leave rubber to fall on the table, it should drop within area where the shadow was dropped previously. There should be point which after falling ( and release of stretching) should remain in precisely the same position.